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Algebraic Equation Manipulation for Electric Circuits
19 questions By Tony R. Kuphaldt
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Question 7 of 19
Manipulate this equation to solve for resistor value R1, given the values of R2 and Rparallel:
$$R_{parallel} = \frac{R_1 R_2}{R_1 + R_2}$$
Then, give an example of a practical situation where you might use this new equation.
Reveal answer$$R_1 = \frac {R_2 R_{parallel}}{R_2 - R_{parallel}}$$
I’ll let you figure out a situation where this equation would be useful!
Notes:This question is really nothing more than an exercise in algebraic manipulation.
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Question 8 of 19
The formula for calculating total resistance of three parallel-connected resistors is as follows:
$$R = \frac {1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}$$
Algebraically manipulate this equation to solve for one of the parallel resistances (R1) in terms of the other two parallel resistances (R2 and R3) and the total resistance (R). In other words, write a formula that solves for R1 in terms of all the other variables.
Reveal answer$$R_1 = \frac {1}{\frac{1}{R}- (\frac{1}{R_2}+\frac{1}{R_3})} \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ R_1 = \frac {1}{\frac{1}{R}-\frac{1}{R_2}-\frac{1}{R_3}}$$
Notes:This question is nothing more than practice algebraically manipulating equations. Ask your students to show you how they solved it, and how the two given answers are equivalent.
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Question 9 of 19
The power dissipation of a transistor is given by the following equation:
$$P = I_C (V_{CE}+ \frac{V_{BE}}{\beta})$$
Manipulate this equation to solve for beta, given all the other variables.
Reveal answer$$\beta = \frac{V_{BE}}{\frac{P}{I_C}- V_{CE}}$$
Notes:Although this question is essentially nothing more than an exercise in algebraic manipulation, it is also a good lead-in to a discussion on the importance of power dissipation as a semiconductor device rating.
High temperature is the bane of most semiconductors, and high temperature is caused by excessive power dissipation. A classic example of this, though a bit dated, is the temperature sensitivity of the original germanium transistors. These devices were extremely sensitive to heat, and would fail rather quickly if allowed to overheat. Solid state design engineers had to be very careful in the techniques they used for transistor circuits to ensure their sensitive germanium transistors would not suffer from “thermal runaway” and destroy themselves.
Silicon is much more forgiving then germanium, but heat is still a problem with these devices. At the time of this writing (2004), there is promising developmental work on silicon carbide and gallium nitride transistor technology, which is able to function under much higher temperatures than silicon.
Question 14, Equation 2 - the ‘+’ symbol is missing between ‘a 3’.
Question 17, right-hand equation - the square-root should only be over the left side of the equation.
Question 19, Hint - the ‘+’ symbol is missing between ‘I 0.005’.
They do seem to be correct in the equivalent questions in the PDF version though.
In the answer to question 8, the second solution has R1 on the right side of the equation when it should be R.